JEE MAIN - Physics (2012 - No. 18)
A wooden wheel of radius $$R$$ is made of two semicircular part (see figure). The two parts are held together by a ring made of a metal strip of cross sectional area $$S$$ and length $$L.$$ $$L$$ is slightly less than $$2\pi R.$$ To fit the ring on the wheel, it is heated so that its temperature rises by $$\Delta T$$ and it just steps over the wheel. As it cools down to surrounding temperature, it process the semicircular parts together. If the coefficient of linear expansion of the metal is $$\alpha $$, and its Young's modulus is $$Y,$$ the force that one part of the wheel applies on the other part is :
$$2\pi SY\alpha \Delta T$$
$$SY\alpha \Delta T$$
$$\pi SY\alpha \Delta T$$
$$2SY\alpha \Delta T$$
Explanation
$$\gamma = {{F/S} \over {\Delta L/L}} \Rightarrow \Delta L = {{FL} \over {SY}}$$
$$\therefore$$ $$L\alpha \Delta T = {{FL} \over {SY}}$$
$$\left[ \, \right.$$ as $${\Delta L = L\alpha \Delta T}$$ $$\left. \, \right]$$
$$\therefore$$ $$F = SY\alpha \Delta T$$
$$\therefore$$ The ring is pressing the wheel from both sides,
$$\therefore$$ $${F_{net}} = 2F = 2YS\alpha \Delta T$$
$$\therefore$$ $$L\alpha \Delta T = {{FL} \over {SY}}$$
$$\left[ \, \right.$$ as $${\Delta L = L\alpha \Delta T}$$ $$\left. \, \right]$$
$$\therefore$$ $$F = SY\alpha \Delta T$$
$$\therefore$$ The ring is pressing the wheel from both sides,
$$\therefore$$ $${F_{net}} = 2F = 2YS\alpha \Delta T$$
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